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4 years ago

it is possible to tell if objects in space are moving closer to us or farther away based on which of the following?

2 answers:

7 0

I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. It is <span>possible to tell if objects in space are moving closer to us or farther away based on several procedures like parallax and standard candles. I hope the answer has come to your help.</span>

Oksana_A [137]4 years ago

5 0

The redshift or blueshift (sorry if its late.)

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A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, $\lambda_1=8\times 10^{-6}\ C$

linear charge density of the other charge distribution, $\lambda_2=-6\times 10^{-6}\ C$

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.r.\epsilon_0}$

where:

$\lambda=$ linear charge density

r = radial distance from the center of wire

$\epsilon_0=$ permittivity of free space

Therefore,

$E_1=E_2$

$\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}$

$\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}$

$\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}$

$x=0.0629\ m$

∴at y=6.29 cm the charge of the two distribution will be equal.

9 0

3 years ago

How does gravity correspond to the bending of the space-time fabric

Klio2033 [76]

The bigger the mass of an astronomical object the bigger the depression it causes in the space-time fabric. Any other astronomical object that gets closer to this depression begins to fall into the depression and hence accelerates closer to the astronomical object causing the depression. This is how gravity is felt.

6 0

4 years ago

A bicyclist starts at point P and travels around a triangular path that takes her through points Q and R before returning to the

REY [17]

Answer:

Displacement by cyclist is zero.

Explanation:

In the given question bicyclist is travelling in a rectangular track having P , Q and R edges.

The bicyclist starts from P and travel through Q and R and returned to P again.

We need to find its displacement.

We know displacement of a body is its difference between its initial position to final position.

Here in the given question the bicyclist returns to P again.

Therefore, total displacement by bicyclist is zero.

Hence, this is the required solution.

4 0

4 years ago

A 0.30 kg softball has a velocity of 15 m/s at an angle of 35 degrees below the horizontal just before making contact with the b

jarptica [38.1K]

Answer:

a) 5.03 kg m/s

b) 10.03 kg m/s

Explanation:

Hi!

Let us consider the origin of coordinates at the pitcher, and pointing directly towards the initial direction of the ball. Therefore, the angle of the velocity with respect to the x axis is -35° (below the horizontal).

The components of the initial momentum are:

px = (0.3 kg)(15 m/s) cos(-35° ) = 4.5 cos(-35° ) kg m/s = 3.69 kg m/s

py = 4.5 sin(-35° ) kg m/s = -2.581 kg m/s

The final momentum will be:

pfx = 0

pfy = - (20 m/s) (0.3 kg) = -6 m/s

And the difference in momentum is:

dpy = pfy - py = -3.419 kg m/s

dpx = pfx - px = -3.69 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 5.03 kg m/s

pfx = -6 kg m/s

pfy = 0

And the difference in momentum is:

dpx = pfx - px = -9.69 kg m/s

dpy = pfy - py = 2.581 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 10.03 kg m/s

6 0

3 years ago

A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

bekas [8.4K]

Answer:

1.98 m/s

Explanation:

To solve this, we would be using the law of conservation of energy, i.e total initial energy is equal to total final energy.

E(i) = E(f)

mgh = ½Iw² + ½mv²

Recall, v = wr, thus, w = v/r

Also, I = ½mr²

I = 0.5 * 5 * 2²

I = 10 kgm²

Remember,

mgh = ½Iw² + ½mv²

Substituting w for v/r, we have

mgh = ½I(v/r)² + ½mv²

Now, putting the values in the equation, we have

5 * 9.8 * 0.3 = ½ * 10 * (v/2)² + ½ * 5 * v²

14.7 = 1.25 v² + 2.5 v²

14.7 = 3.75 v²

v² = 14.7/3.75

v² = 3.92

v = √3.92

v = 1.98 m/s

Thus, the speed is 1.98 m/s

5 0

3 years ago